1.
Present an NFA having at most five states and eight transitions
that accepts the language
In other words, the language contains precisely those strings over the alphabet {a,b} having abba as a substring.
(A transition labeled by both a and b counts as two transitions.)
2. Present an NFA having at most six states that accepts the language
In other words, the language contains precisely those strings over the alphabet {a,b} having either bab as a substring or baaa as a suffix.
3. Present an NFA having at most four states that accepts the language
In other words, the language contains precisely those strings over the alphabet {a,b,c} in which at least one among the symbols a, b, and c fails to appear.
4. Present an NFA having at most seven states (note that only one accepting/final state is necessary) that accepts the language
In other words, the language contains precisely those strings over {a,b} in which either a is the fourth-to-last symbol or b is the third-to-last symbol.
5. Present an NFA having four states and at most two transitions labeled b that accepts the language
In other words, the language contains precisely those strings over {a,b} that can be "factored" into a sequence of zero or more substrings, where each factor is either ab, aab, or aba. An example of such a string (in which vertical bars have been inserted in order to show the boundaries between adjacent factors) is
In each of the last two problems, you are given an NFA M and asked to present the (equivalent) DFA that results from applying to M the "subset construction" algorithm.
Present your answers in tabular form (such as how the NFAs are described here) rather than as (or in addition to, if you want) a transition graph. Don't forget to identify the initial state and the accepting/final states.
Your DFA's state names should make obvious which subset of the NFA's state set each one corresponds to. For example, q{0,2,3} could be the name of the state corresponding to {q0, q2, q3}.
6. The NFA is M = (Q, {a,b}, δ, q0, {q3}), where Q = {qi : 0≤i<4} and δ is as described in the table below:
| a | b | |
|---|---|---|
| q0 | {q1,q2} | {q1} |
| q1 | {q0} | {q1} |
| q2 | {q3} | {q3} |
| q3 | {q1} | ∅ |
The resulting DFA should have eight states.
7. The NFA is M = (Q, {a,b}, δ, q0, {q1}), where Q = {qi : 0≤i<4} and δ is as described in the table below:
| a | b | λ | |
|---|---|---|---|
| q0 | {q1} | ∅ | ∅ |
| q1 | {q2} | ∅ | {q3} |
| q2 | ∅ | {q3} | ∅ |
| q3 | {q1} | {q2} | ∅ |
The resulting DFA should have seven states.